
#1
Mar2411, 03:38 PM

P: 196

1. The problem statement, all variables and given/known data
Find voltage between node 1 and node 2 2. Relevant equations What is the algorithm? 3. The attempt at a solution R1=4 Ohms R2=5 Ohms R3=8 Ohms R4=3 Ohms R5=6 Ohms R6=7 Ohms Eэ=20 V E3=17 V E5=15 V 



#2
Mar2411, 03:47 PM

P: 1,135

use balanced wheat stone bridge thing




#3
Mar2411, 03:55 PM

P: 196

Do I need second Kirhgoff rules' equations for every loop?




#4
Mar2411, 04:01 PM

P: 1,135

find voltage between two nodes
try "balanced wheat stone bridge"
you wont need Kirhgoff rules here:http://en.wikipedia.org/wiki/Wheatstone_bridge 



#5
Mar2411, 04:04 PM

Sci Advisor
HW Helper
Thanks
P: 26,167

hi builder_user!
'fraid so! 



#6
Mar2411, 04:20 PM

P: 196

What's the next step after Kirchgoff's rules?What do I need to find?I know current in every branch.




#7
Mar2411, 04:29 PM

Mentor
P: 11,441

You want the voltage between nodes 1 and 2. Use the current you found for the first loop (leftmost) using Kirchhoff to determine the voltage across R1. Then do the obvious voltage sum.




#8
Mar2411, 04:35 PM

P: 196

i1=3.7A
U=Eэi1*R1=5.2V? And the final result  20(Eэ)+5.2(U)=25.2V? 



#9
Mar2411, 04:49 PM

Mentor
P: 11,441

That's not the value I get for i1. Better check your derivation.
EDIT: My error. Redoing my sums I see that i1 is indeed about 3.7A. 



#10
Mar2411, 05:14 PM

P: 196





#11
Mar2411, 05:17 PM

Mentor
P: 11,441





#12
Mar2411, 05:23 PM

P: 196

but i's only I1.
Or R1*I1  is the result? 



#13
Mar2411, 05:24 PM

Mentor
P: 11,441

Very sorry. I see that your figure of 3.7A for i1 and about 5.2V for the voltage between nodes 1 and 2 look okay.




#14
Mar2411, 05:36 PM

P: 196





#15
Mar2411, 05:46 PM

Mentor
P: 11,441

Again, sorry about that.
15 pages of A4? That seems like a lot for just three loops. What method are you using to solve the equations? 



#16
Mar2411, 05:50 PM

P: 196

I got this result(3.7) by different methods.So if it's wrong  all my work is wrong 



#17
Mar2411, 06:04 PM

Mentor
P: 11,441

I see, so this is just one exercise for the given schematic.
It seems to me that a KVL loop approach is the most straight forward for this particular question, since you really only need to solve for the current in the first loop. Have you learned the method to directly write (by inspection) the matrix form of the equations? That allows you solve for the currents using a matrix method (such as Cramer's Rule), which takes only a few lines. 



#18
Mar2411, 06:12 PM

P: 196

if it's cramer's method...
I know this method but I only use it in programming.I solve all equtations in MathCAd.And then I show result without numbers and then the final result.The result without numbers is very big sometimes. 


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